Talk:Surgery obstruction map I (Ex)
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& = \langle L(\eta)^{-1},[X]\rangle - \langle L(TX),[X]\rangle | & = \langle L(\eta)^{-1},[X]\rangle - \langle L(TX),[X]\rangle | ||
\end{array}$$ | \end{array}$$ | ||
− | by the Hirzebruch signature theorem and | + | by the Hirzebruch signature theorem and several properties of the $L$-genus. In particular the surgery obstruction depends only on the bundle over $X$. |
Now $(\xi\oplus\xi,\phi * \phi)$ is the sum of $(\xi,\phi)$ and $(\xi,\phi)$ in $\mathcal N (X)=[X,G/O]$ with respect to the Whitney sum. | Now $(\xi\oplus\xi,\phi * \phi)$ is the sum of $(\xi,\phi)$ and $(\xi,\phi)$ in $\mathcal N (X)=[X,G/O]$ with respect to the Whitney sum. | ||
Moreover | Moreover |
Revision as of 18:27, 31 May 2012
If is a manifold, then the normal map gives the base point of . An element of is given by a bundle together with a fiber homotopy trivialization . Under the isomorphism , the pair corresponds to a normal map covered by . Assume that the dimension is and that is simply connected. Then the surgery obstruction of a normal map covered by equals
by the Hirzebruch signature theorem and several properties of the -genus. In particular the surgery obstruction depends only on the bundle over . Now is the sum of and in with respect to the Whitney sum. Moreover
If this is non-zero, then the surgery obstruction is not a group homomorphism with respect to the Whitney sum.
As an example take :
There are fiber homotopically trivial bundles on corresponding to classes in which restrict to any given class in , as follows from the Puppe sequence with . From another exercise we know that on we have such vector bundles with first Pontryagin class times the generator of . This means that on we have a vector bundle with whose sphere bundle is fiber homotopically trivial, by a fiber homotopy equivalence . We compute
where the constant can be computed from the L-genus to be .
So the surgery obstruction is not a group homomorphism with respect to the Whitney sum.